Codeforce 580C. Kefa and Park-优快云博客

本文介绍了一个关于寻找满足特定条件路径的算法问题。主人公Kefa想要从家出发到公园内的餐厅用餐，但害怕经过太多连续有猫的节点。通过深度优先搜索（DFS），文章提供了一种解决方案来计算满足条件的餐厅数量。

Kefa decided to celebrate his first big salary by going to the restaurant.

He lives by an unusual park. The park is a rooted tree consisting of n vertices with the root at vertex 1. Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices with cats in them.

The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than m consecutive vertices with cats.

Your task is to help Kefa count the number of restaurants where he can go.

Input

The first line contains two integers, n and m (2 ≤ n ≤ 105, 1 ≤ m ≤ n) — the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa.

The second line contains n integers a1, a2, ..., an, where each ai either equals to 0 (then vertex i has no cat), or equals to 1 (then vertex i has a cat).

Next n - 1 lines contains the edges of the tree in the format "xi yi" (without the quotes) (1 ≤ xi, yi ≤ n, xi ≠ yi), where xi and yi are the vertices of the tree, connected by an edge.

It is guaranteed that the given set of edges specifies a tree.

Output

A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most m consecutive vertices with cats.

简单的dfs，注意对边的处理和叶子节点的判断

#include <cstdio>
#include <cctype>
#include <stdlib.h>
#include <iostream>
#include <cmath>
#include <iomanip>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <map>

using namespace std;
typedef long long LL;
int cat[100005];
vector<int> t[100005];
int n,m;
int total = 0;
int visited[100005] = {0};
void search(int now,int catnow){
  visited[now] = 1;
  // cout << now << endl;
  if (cat[now] == 1){
    catnow ++;
  }
  else {
    catnow = 0;
  }
  if (catnow > m){
    return;
  }
  int empty = 1;
  for (int i=0;i<t[now].size();i++){
      if (!visited[t[now][i]]){
        empty = 0; break;
      }
  }
  if (empty){
    total ++; return;
  }
  for (int i=0;i<t[now].size();i++){
    int after = t[now][i];
    if (visited[after]) continue;
    // cout << now << " " << after <<endl;
    search(after,catnow);
  }
}

int main()
{
  // freopen("test.in","r",stdin);

  cin >> n >> m;
  for (int i=1;i<=n;i++){
    cin >> cat[i];
  }
  for (int i=1;i<n;i++){
    int a,b;
    cin >> a >> b;
    t[a].push_back(b);
    t[b].push_back(a);
  }
  search(1,0);
  cout << total;
  return 0;
}